Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.

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Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.

Mathematics
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Okay, you can find the volume first since you have h and r , so : \[V = \pi r^2 h\] \[= ( 3)^2 (6) \pi\] = 169.65 ft^3
sstarica, isn't this the same thing we were tackling
now find V' which is :\[V' = 2\pi rr'h'\]

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Other answers:

not really @iam , somehow similar :)
now we have: h' = 0.2 ft/sec we need to find r'
do i just plug in
yes :) but we need to find r' now
\[V' = 2 \pi rr' h'\] we have h' , but we need to find r' to calculate V' = rate of volume increase
how do i do that? :(
hold on :)
thx
you need to find a relationship between r and h so r/h = r'/h'
thats where im having problem for all the problems
cross mulitply and you'll get: r' = rh'/r
sorry! wait you'll get: r' = rh'/h :)
are you with me soo far?
yes :)
now plug : r =3 ft h = 6ft h' = 0.2ft/sec r' = (3)(0.2)/6 = 0.1 ft/sec
we have found r' now, let's go back to V'\[V' = 2 \pi rr'h'\] so, V' = 2(3)(0.1)(0.2) pi = 0.377 ft^3/sec
Correct me if I'm wrong please
OMG this makes so much sense now
but the website marked it wrong :(
why?
I guess there's a mistake, but that's the idea, maybe in calculation?
ok
did you get the idea?
one qquestion
sure :)
where did u get \[V \prime = 2\pi r r \prime h \prime \]
you derive with respect to r and h :)
you want to know how is the volume increasing when both the radius and height is changing, that's the story of the related rated problem
oooooh...im sorry can u show me one more problem
hmmmm, let's say we have a circle of radius = 2 m, and the radius is decreasing at 0.5 m/sec. How much will the circle's area increase ?
I'll give you advise, whenever you see ( m/sec) or anything with respect to time it's something' which in this case is the r' = 0.5m/sec
so what is the given?
we have : - R = 2 m - R' = 0.5 m/sec and we know the formula of the Area of the circle which is: A = pi r^2, right?
uhhuh
Now , you want to know how much is the area increasing with respect to R, since R is increasing :) so you find the derivative of it which will be : A' = 2 pi rr'
so when exactly do i start pluging in numbers?
right now, after you find the derived formula, start pluggingthe numbers :)
okay ...these are pretty simple compared to the ones i have on hw :'(
Like this problem, they want to know how much is the area increasing , we got the formula so the answer is: A' = 2(2)(0.5) pi = 2 pi = 2.68 m^2/sec :) So the area is increasing at 2.68 m^2/sec
Lol, when you understand the small idea, you'll be able to figure out the bigger idea :)
did you understand my problem?
yes but how would i approach towards a hemisphere problem
think about it, write down the given on one side, and the RTF(required to find) on the other side, then SKETCH! To test your understanding , you've got to sketch what they give you to be fully understand what they want :)
to fully*
1) write the given and what you need to find 2) sketch 3) write down the general formula that you're going to use :)
^_^
ok :)
good luck ! You're a smart student, you can figure it out :)
you got the small idea, you'll be able to solve any bigger ideas now . They are not even big, they are small ideas that got stretched ^_^, don't let the problem boss u, YOU ARE THE BOSS!
lol..thx
np :) Give it a try once more. ^_^

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